Multiply the following complex numbers: $({1}) \cdot ({3-4i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({1}) \cdot ({3-4i}) = $ $ ({1} \cdot {3}) + ({1} \cdot {-4}i) + ({0}i \cdot {3}) + ({0}i \cdot {-4}i) $ Then simplify the terms: $ (3) + (-4i) + (0i) + (0 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 3 + (-4 + 0)i + 0i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 3 + (-4 + 0)i - 0 $ The result is simplified: $ (3 - 0) + (-4i) = 3-4i $